1-find-the-locus-z-z-1-2-meets-the-positive-real-axis-2-On-a-single-Argand-diagram-sketch-the-loci-z-z-1-2-arg-z-z-2-pi-4-

Question Number 112454 by bemath last updated on 08/Sep/20

(1) find the locus ∣ z - z_{1} ∣ = 2 meets

the positive real axis

(2) On a single Argand diagram, sketch

the loci \to {\begin{cases} ∣ z - z_{1} ∣= 2 \\ \arg (z - z_{2}) = \frac{π}{4} \end{cases}

Commented by MJS_new last updated on 08/Sep/20

(2) we cannot scetch the loci if we {don}^{'} t know

any coordinates .

∣ z - z_{1} ∣= 2 is the equations of all circles with

radius 2

\arg (z - z_{2}) = \frac{π}{4} is any straight line with slope 45 °

let z - z_{2} = r e^{i θ} \Rightarrow \arg r e^{i θ} = θ \Rightarrow θ = \frac{π}{4}

\arg ((x - p) + (y - q) i) = \frac{π}{4}

leads to y = x - p + q

Answered by MJS_new last updated on 08/Sep/20

(1)

assuming z_{1} is given and z is variable we have

z_{1} = p + q i \land z = x + y i

∣ z - z_{1} ∣= 2

∣ (x - p) + (y - q) i ∣= 2

\sqrt{{(x - p)}^{2} + {(y - q)}^{2}} = 2

both sides are ⩾ 0 \Rightarrow we are allowed to square

{(x - p)}^{2} + {(y - q)}^{2} = 4

which is a circle with center (\begin{matrix} p \\ q \end{matrix}) and radius 4

if it meets the positive real axis depends on

the values if p and q

- 2 ⩽ q ⩽ 2 \Rightarrow it meets the real axis

p > 2 \Rightarrow it meets the positive real axis

we find the intersection point (s) putting

y = 0

{(x - p)}^{2} + q^{2} = 4

\Rightarrow x = p \pm \sqrt{4 - q^{2}} with p > 2 \land - 2 ⩽ q ⩽ 2

Commented by bemath last updated on 08/Sep/20

thank you sir

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